# Engineering Hydrology Questions and Answers – Synthetic Unit Hydrograph – Set 5

This set of Engineering Hydrology Multiple Choice Questions & Answers (MCQs) focuses on “Synthetic Unit Hydrograph – Set 5”.

1. What is the relationship between the lag time (t) and time of concentration (tc) for a basin as per SCS?
a) t = 0.6*tc
b) t = 1.67*tc
c) tc = 0.6*t
d) tc = 1.67*t

Explanation: The US Soil Conservation Services have conducted hydrologic studies on many small sized rural catchment areas and found that the lag time is approximately equal to 60% of the time of concentration.

2. In a SCS triangular unit hydrograph, what is the product of the peak discharge and time to peak for a catchment of area 1450 hectares?
a) 0.3
b) 4.8
c) 30
d) 48

Explanation: Let Tp be the time to peak (hr), Tb be the base time (hr) and Qp be the peak discharge (m3/s) of the triangular hydrograph. Let A be the area of the catchment in hectares. Now, the area under the hydrograph is equal to the volume of runoff of 1 cm depth.
$$\frac{1}{2}*Q_p(\frac{m^3}{s})*T_b$$(hr)=1 (cm)*A (ha)
→ $$\frac{1}{2}*Q_p(\frac{m^3}{s})$$*3600*Tb (secs)=10-2(m)*A*104 (m2)
→ Qp*Tb=0.0556*A
→ Qp*(2.67*Tp)=0.0556*A
→ Qp*Tp=0.0208*A=0.0208*1450=30.16≅30

3. Find the time to peak (in minutes) for a 20 minute SCS triangular hydrograph and for a time of concentration of 45 minutes?
a) 37
b) 43
c) 45
d) 51

Explanation: Given D = 20 min = 0.33 hr and tc = 45 min = 0.75 hr
So lag time, t=0.6*tc = 0.6*0.75 = 0.45 hr
Therefore, time to peak,
Tp=$$\frac{D}{2}$$+t=$$\frac{0.33}{2}$$+0.45=0.615 hr=0.615*60 minutes=36.9 minutes≅37 minutes

4. What will be the ratio of base time to lag time for a 60 minute SCS triangular unit hydrograph for a watershed of 10km2 area and time of concentration of 80 minutes?
a) 1.63
b) 2.67
c) 4.34
d) 5.44

Explanation: Given D = 60 min = 1 hr, tc = 80 min = 1.33 hr, A = 10 km2
Lag time, t = 0.6*tc = 0.6*1.33 = 0.798 hr
Time to peak, tp = (D/2) + t = 0.5 + 0.798 = 1.298 hr
Base time, tb = 2.67*tp = 2.67*1.298 = 3.466 hr
Therefore, required ratio = $$\frac{t_b}{t} = \frac{3.466}{0.798}$$ = 4.34

5. Which of the following is not a part of the long term approach adopted by CWC for estimating design flood discharges applicable to catchments ranging from 25-1000 hectares in area?
a) Calculation of weighted mean slope of the catchment
b) Dividing the country into hydrologically homogeneous subzones
c) Developing synthetic unit hydrographs for a catchment
d) Documenting the details of procedure, relations and limitations of methods used

Explanation: The Central Water Commission of India adopted two approaches for estimation of design flood peaks, i.e., long term and short term. The short term approach is a quick method using empirical equations which requires the calculation of weighted mean slope.

6. For a catchment with a weight mean slope (S) of 0.005 and area A (in km2), what is the correction formula for finding the peak discharge of a unit hydrograph as per CWC short term plan?
a) 1.79 * A3/4
b) 37.4 * A3/4
c) 1.79 * A3/4 * S2/3
d) 37.4 * A3/4 * S2/3

Explanation: The CWC short term plan gives two formulae for calculating peak discharge depending on whether the weighted mean slope of the catchment is more than or less than 0.0028. The peak discharge,
Qp = $$\begin{cases}1.79 * A^{\frac{3}{4}} & for \, S > 0.0028\\37.4 * A^{\frac{3}{4}} * S^{\frac{2}{3}} & for \, S < 0.0028\end{cases}$$

7. A 1-hr unit hydrograph for a catchment of area 14 km2 has a peak discharge of 13 m3/s. What is the lag time (in minutes) for this unit hydrograph as per CWC?
a) 17
b) 55
c) 87
d) 100

Explanation: The lag time of a 1-hr unit hydrograph according to CWC short term plan is given by,
t1-hr=$$\frac{1.56}{(\frac{Q_p}{A})^{0.9}} = \frac{1.56}{(\frac{13}{14})^{0.9}}$$ = 1.67 hours=100 mins

8. How many subzones has India been divided into, to carry out the CWC long term plan for estimating design flood peaks?
a) 23
b) 26
c) 28
d) 30

Explanation: Under the long term plan, India has been divided into 26 hydrometeorologically homogeneous subregions. A regional synthetic hydrograph has been derived for each of these subzones.

Sanfoundry Global Education & Learning Series – Engineering Hydrology.